Question

Prove that the sum of the squares on the sides of a rhombus is equal to the
sum of squares on its diagonals.
olution. ​

Answer 1

Answer :-

Step-by-step explanation :-

Given :- A rhombus ABCD whose diagonals AC and BD intersect at O.

To Prove :- ( AB² + BC² + CD² + DA² ) = ( AC² + BD² ).

Proof :-

➡ We know that the diagonals of a rhombus bisect each other at right angles.

=>∠AOB = ∠BOC = ∠COD = ∠DOA = 90°,

$OA = \frac{1}{2} AC and OB = \frac{1}{2}BD$

From right ∆AOB , we have

AB² = OA² + OB² [ by Pythagoras' theorem ]

AB² = $(\frac{1}{2} AC)^{2} + (\frac{1}{2}BD)^{2}$

AB² =$\frac{1}{4} (AC^{2} + BD^{2} )$

==> 4AB² = ( AC² + BD² ) .

==> AB² + AB² + AB² + AB² = ( AC² + BD² ) .

•°• AB² + BC² + CD² + DA² = ( AC² + BD² ) .

[ In a rhombus , all sides are equal ] .

Hence, it is proved.

Thank U : )

Answer 2

Question:

Prove that the sum of the squares of the sides of a rhombus is equal to the sum of square of its diagonals

Answer:

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• A rhombus ABCD whose diagonals are perpendicular bisectors

$\Large{\underline{\underline{\bf{To\:Prove:}}}}$

• AB² + BC² + CD² + AD² = BD² + AC²

$\Large{\underline{\underline{\bf{Proof:}}}}$

→ Consider Δ ODC

→ By Phythagoras theorem,

  CD² = OD² + OC²

→ Here OD = 1/2 BD and OC = 1/2 AC

→ Substitute these in the above equation,

  CD² = (1/2 BD)² + (1/2 AC)²

→ Expanding it,

  CD² = 1/4 BD² + 1/4 AC²

→ Multiply the whole equation with 4

  4 CD² = BD² + AC²

  CD² + CD² + CD² + CD² = BD² + AC²

  AB² + BC² + CD² + AD² = BD² + AC²

  Since AB = BC = CD = DA in a rhombus

→ Hence proved.

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ In a rhombus, all sides are equal

→ The diagonals of a rhombus are perpendicular bisectors.

→ The opposite sides are parallel to each other